In a second-order reaction, if first-order is observed for both the r…

In a second-order reaction, where both reactants A and B exhibit first-order behavior, the rate of the reaction is proportional to the product of the concentrations of A and B.

Let's examine the reactant mixtures provided in the options:

(1) 0.1 mol of A and 0.1 mol of B in 0.2 litre solvent
Product of concentrations: \( (0.1 \, \text{mol}) \times (0.1 \, \text{mol}) = 0.01 \, \text{mol}^2 \)

(2) 1.0 mol of A and 1.0 mol of B in one litre solvent
Product of concentrations: \( (1.0 \, \text{mol}) \times (1.0 \, \text{mol}) = 1.0 \, \text{mol}^2 \)

(3) 0.2 mol of A and 0.2 mol of B in 0.1 litre solvent
Product of concentrations: \( (0.2 \, \text{mol}) \times (0.2 \, \text{mol}) = 0.04 \, \text{mol}^2 \)

(4) 0.1 mol of A and 0.1 mol of B in 0.1 litre solvent
Product of concentrations: \( (0.1 \, \text{mol}) \times (0.1 \, \text{mol}) = 0.01 \, \text{mol}^2 \)

Comparing the product of concentrations, we can see that option (3) with \(0.04 mol^2\) provides the highest value. Therefore, the reactant mixture of 0.2 mol of A and 0.2 mol of B in 0.1 litre of solvent will provide the highest initial rate.

Therefore, the correct answer is \((3) \, 0.2 \, \text{mol} \, \text{of} \, A \, \text{and} \, 0.2 \, \text{mol} \, \text{of} \, B \, \text{in} \, 0.1 \, \text{litre} \, \text{solvent}\).